jestingrabbit wrote:Just to lay out some of the more obvious things that we do actually know. We definitely have that tan is monotonically increasing and unbounded on the relevant interval. Further, I believe that the tan angle sum is just a consequence of basic trig stuff, no limits needed.

Yes, both of those things can be shown without heavy machinery.

jestingrabbit wrote:So, from this, we can conclude that there is a greatest t such that t = tan(t). I think rubbing these two things together in the right way should lead to a contradiction of some sort, but its not going to be elegant, that is for sure.

In order to conclude from those premises that there is a greatest fixed point of tan(x) on [0, π/2), we would also need to prove that “π/2” (which I am using to denote the arc length of a quarter circle) is itself finite. Put another way, we need an upper bound on the circumference of a circle first.

It is easy enough to put a lower bound on circumference, because a straight line is the shortest path between endpoints. And it is easy enough to put both upper and lower bounds on area, because containment puts a partial order on area. But I do not see an obvious way to put an upper bound on the length of a curve.

Hmm, if we assume that a circle has finite circumference which grows with radius (we don’t even need that it’s proportional, just monotone!), then we can develop some

extremely rudimentary “calculus”. We’ll need the fact that the area of a circle is proportional to its radius squared, which is easy enough to show with triangles. We also need one more fact, which I will call the…

Hat brim lemma: In a circular top hat of height h, if the brim extends out from the hat by that same distance h, then the brim has larger area than the vertical “stovepipe” wall of the hat.

Proof: Remove the flat top of the hat and unroll the wall-and-brim. The outer part of the brim shrinks during this process, while the vertical wall keeps its area. When they are straightened, both pieces become rectangles of equal width (h) and length (the inner circumference), so they end up with equal areas. The brim shrank and the wall didn’t, so the brim was larger at the outset.

∎

Now we can compare a circle of radius r with one of radius (r+h). Call their areas A and B, respectively, and let c be the circumference of the smaller circle. By the hat brim lemma we have B > A+hc. But area is proportional to radius squared, so we also know that A/r

^{2} = B/(r+h)

^{2}, which yields B = A(r+h)

^{2}/r

^{2}.

Plugging that into the previous inequality gives A(r+h)

^{2}/r

^{2} > A+hc. Some algebra (ugh, is there a good way to show this geometrically?) yields A(2hr+h

^{2})/r

^{2} > hc, and then A(2 + h/r) > rc.

For a fixed radius r, we can take h as small as we like and the inequality still holds. Since A is a constant, it follows that 2A cannot be less than rc (otherwise a small enough h would break the inequality), so we must have A ≥ rc/2.

The same argument holds for any sector of a circle (just slice the top hat accordingly) which means A

_{sector} ≥ rs/2. We could do the same thing but with the wall outside the brim (so the brim expands) to show that A ≤ rx/2, and thus A = rx/2 as well as A

_{sector} = rs/2, but for now just having the upper bound suffices.

Looking at the unit circle (in the first quadrant), by containment of areas we have tan(x)/2 > A

_{sector}. With s=x and r=1 we get A

_{sector} ≥ x/2, hence tan(x)/2 > x/2 or simply tan(x) > x. Taking reciprocals and multiplying by sin(x) gives cos(x) < sin(x)/x. The upper bound of 1 is easy to get, as is the fact that cos(x) → 1. Symmetry gives us the same results in the third quadrant, which shows that sin(x)/x → 1 if we accept the hat brim lemma (including its assumption that circles have finite circumference).

Of course, my justification for the hat brim lemma is not exactly rigorous. It appeals to intuition, and is boosted by the fact that circles are radially-symmetric so the same contraction of the brim happens in all directions, but it still talks about bending surfaces in three dimensions.

Can we provide a stronger (preferably two-dimensional) foundation for the hat-brim lemma, which shows that the area of an annulus is at least its radial width times its inner circumference?